Question

Two simple pendulums of lengths \(L_1\) and \(L_2\) have periodic time \(T_1\) and \(T_2\) respectively \((T_1 > T_2)\). The time period of the pendulum of length \((L_1-L_2)\) is
\([(L_1-L_2) > 60\text{ cm}]\)

• A. \(\sqrt{T_1^2+T_2^2}\)
• B. \(\sqrt{T_1^2-T_2^2}\)
• C. \(T_1+T_2\)
• D. \(T_1-T_2\)

Hint:

Use \(T^2\propto L\) and subtract.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
For a simple pendulum, \(T=2\pi\sqrt{L/g}\), so \(T^2=\dfrac{4\pi^2}{g}L\).

Step 2: Key Formula or Approach
\(T_1^2=\dfrac{4\pi^2L_1}{g}\) and \(T_2^2=\dfrac{4\pi^2L_2}{g}\).

Step 3: Detailed Explanation
\[ T_1^2-T_2^2=\frac{4\pi^2}{g}(L_1-L_2)=T^2 \]
\[ T=\sqrt{T_1^2-T_2^2} \]

Final Answer:
The time period is \(\sqrt{T_1^2-T_2^2}\), option (B). \[ \boxed{\sqrt{T_1^2-T_2^2}\ \text{(B)}} \]